Disclaimer: The following material is being kept online for archival purposes.

Although accurate at the time of publication, it is no longer being updated. The page may contain broken links or outdated information, and parts may not function in current web browsers.

Get a Straight Answer

Please note!

Listed below are questions submitted by users of "From Stargazers to Starships" and the answers given to them. This is just a selection--of the many questions that arrive, only a few are listed. The ones included below are either of the sort that keeps coming up again and again, or else the answers make a special point, often going into details which might interest many users.

For a complete list, including later questions not listed below, click here. You may also link from here to a listing of questions arranged by topic.

Part of the problem was that Goddard preferred to work
alone, while the Caltech people brought in bright students
and had much better engineering support.

I never heard about the Germans spying on Goddard, and it seems
very unlikely. They too had much better engineering support
and took Goddard's ideas--DeLaval nozzle, liquid fuel,
using the fuel to cool the engine, steering vanes in the
exhaust etc.--and developed them beyond what Goddard himself
was able to do.

A similar thing happened in WW-I. The Wright brothers
invented the airplane in 1903, but the Europeans took their
work and expanded it greatly, so that the German, British
French and even Russian airplanes in that war were far
superior to the ones America produced. After America entered
the war, its pilots all flew British and French machines.

A ball of mass M1 moving with velocity – V1, against...
A paddle of mass M2, moving with velocity V2

Afterwards we have

A ball of mass M1 moving at velocity +W1
A paddle of mass M2 moving with velocity W2

We assume the paddle is much more massive--M2 >> M1 (actually, it is mostly the mass of the hand behind the paddle), so that V2 and W2 are almost the same (=the impact does not slow the paddle by any great amount).

Conservation of momentum:

M2V2 – M1V1 = M2W2 + M1W1
(1)

Conservation of energy (we assume the encounter is perfectly elastic--
approximate for the ping-pong ball, very well observed by gravity-assist maneuvers of spacecraft around planets):

M2V22/2 + M1V12/2 = M2W22/2 + M1W12/2

multiply by 2:

M2V22 + M1V12 = M2W22 + M1W12 (2)

In both numbered equations we collect all M2 terms on the left and all M1 terms on the right:

M2(V2 - W2) = M1(W1 + V1)
(3)

M2[V22 – W22] = M1[W12 – V12] (4)

By a well known factoring identity, for any two numbers A and B

A2 – B2 = (A + B)(A – B)

so (4) becomes

M2(V2 – W2)(V2 + W2) = M1(W1 – V1)(W1 + V1) (5)

If we divide equals by equals, what remains is still a valid equality.
So let the left side of (5) be divided by the left of (3), the the right
side of (5) by the right of (3):

I expect the discrepancy stems from the fact that the model used in
From Stargazers to Starships is based on the ping-pong paddle
example. The key difference is that the force exerted by a ping-pong
paddle on a ball is repulsive, whereas gravity is attractive. Thus the
numbers are the same but the sign is reversed.

By the way, I did find the analogy to the Pelton turbine very interesting.
Thank you again for a very informative web site!

Reply

I believe that the ping-pong analogy is still valid, because it can be
reduced to simple arguments of the conservation of momentum and energy,
which should hold equally in a planetary-assist maneuver. Some other
correspondent questioned this result, and as a result, you can find that
calculation in item #9 of the question-and-answer section of
"Stargazers," linked at the end of the home page [the item preceding this one].

What seems to confuse the issue is the following. A spacecraft would
get its biggest boost if it approached head-on, made a hairpin turn
around the rear of the moving planet and returned along a path 180
degrees from its first approach (that would be the ping-pong analogy).
Viewing the encounter from far north, if we put the moving planet at the
center of a clock dial with the Sunís direction at 12 o'clock, we would
see the planet moving towards 3 o'clock, so our satellite has to
approach from that direction and return to it again.

When the Voyager and Pioneer spacecraft approached Jupiter and Saturn,
however, they were coming from the Earth, which is roughly in the same
direction as the Sun; in any case, their initial orbital velocity, which was essentially that of the Earth, which moves in the same direction as other planets', did not allow a head-on approach. Instead, they entered around 12 o'clock on the dial. They still rounded the night side and exited around 3 o'clock, which gave them an apprecible boost, though perhaps not the biggest one possible.

I have some old issues of "Science" on these events and in the one of
the Pioneer 10 fly by, for instance (page 304, 25 January 1974), the
satellite enters at 1 o'clock and leaves a bit after 3 o'clock. For the
Voyager 1 fly-by of Saturn (p. 160, April 10, 1981), entrance is around
11:30 and exit around 4:30 on the same dial.

You are right, of course, in that the force on the ping-pong ball is
repulsive while the planet's gravity attracts the spacecraft. However,
the strongest attraction occurs when the spacecraft is at its closest
approach, on the night side, and its direction then is along the
velocity of the planet, the same direction as the force exerted in the
ping-pong analogy.

With all this, I am grateful for your message. It again shows that at
least some users go into the details of "Stargazers". Quite a few errors
were caught only thanks to people like yourself who checked out such
details.

Take a sheet of paper, put on it a small circle--that is the Sun viewed from far north of it, or rather, it is a circle in the corona, some level above the Sun, where the solar wind begins. On this scale, let's say the solar wind moves one inch (1") per day (or if you wish, 2 cm). Draw from the center 6 or 7 radial rays 13.3 degrees apart. Mark as "P" the point where the first ray--the one furthest clockwise--cuts the circle. We look at 6 ions located at P, and
therefore presumably on the same field line--let's number them 1, 2...6.
We have advance information that 1 will be released into the solar wind
today, 2, tomorrow, 3 the day after, and so on. Mark P with 1--that is
where ion no. 1 is today.

Next day, P is on the second ray. Point 1 has moved 1" outward, radially, and Point 2 is at the base of the new ray, ready to go. Next day: Point 1
is now 2" out on the first ray, point 2 is 1" out on the 2nd, point 3 at the base of the 3rd, ready to move. And so on.

Five days later, 1 is 5" out on the first ray, 2 is 4" out on the 2nd
3 is 3" out on the 3rd, etc., and 6 is at the base of the 6th ray.
However, all these points started on the same field line, so they are
still strung out along one line. CONNECT THE DOTS marking the outermost ions on the 6th day and you have a spiral line of the interplanetary field: if the ions started on the same line, they must still be on one.

The solar wind in all this has moved radially. But now and then the
sun releases bursts of high energy particles, say from flares. The
energy of these particles may be high enough to endanger astronauts
in interplanetary space--but their density is very low, so their beta is
also low. THEY therefore are guided by the magnetic field lines (rather
than deforming them to their own flow), and therefore they move spirally.

The solar wind takes about 5 days to cover 1 AU. Therefore, if the Earth
is to receive particles guided by an interplanetary field line when it
is on the first ray, the emission has to be at the base of ray 6--that
is, near the western limb. The high-energy particles take only an hour or
so to arrive, depending on their energy of course.

I was recently looking at the webpage "Seasons of the Year" and I read about what would happen if the earth's axis were perpendicular to the ecliptic. I was just wondering if you could give me some insight on what would happen if the ecliptic was inclined at a 90-degree angle with respect to the celestial equator? Would this mean that earth's orbit would
travel along this "new ecliptic" while the north and south poles are
travelling along this "new ecliptic"?

Reply

The hypothetical case you describe does in fact exist: for some
unknown reason, the spin axis of the planet Uranus is almost exactly
in the ecliptic.

That means that at some time one pole (let me call it the north
pole, even though that "north" direction is almost perpendicular
to the northward direction from Earth) points at the Sun. Then the
northern hemisphere is in constant light and the other one in
constant darkness. Half an orbit later--42 years or so--the roles
are reversed. And halfway between those times, the planetary rotation
axis is perpendicular to the Sun's direction, making day and night
alternate in a way similar to what the Earth experiences at equinox.

I leave it as an exercise to you to figure out whether Uranus ever
receives sunlight the way Earth does at solstice.